Consider the equilibrium reaction 2N2(g) + O2(g) = 2N2O(g) In a particular experiment, the equilibrium concentration of N2 is 0.048 M, of O2, 0.093 M, and of N20, 6.55 x 10-21 M. What is the value of the equilibrium constant K,? а. 1.5 х 10-18 b. 2.0 х 1037 с. 2.2 х 10-36 d. 3.1 x 10-17 е. 5.0 х 10%6

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**Chemical Equilibrium Constant Calculation**

Consider the equilibrium reaction:

\[ 2N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g) \]

In a particular experiment, the equilibrium concentration of N₂ is 0.048 M, of O₂ is 0.093 M, and of N₂O is 6.55 x 10⁻²¹ M. What is the value of the equilibrium constant \( K_c \)?

**Possible answers**:

a. \( 1.5 \times 10^{-18} \)

b. \( 2.0 \times 10^{-37} \)

c. \( 2.2 \times 10^{-36} \)

d. \( 3.1 \times 10^{-17} \)

e. \( 5.0 \times 10^{36} \)

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Interpretation:

To determine the equilibrium constant \( K_c \), use the expression for the equilibrium constant \( K_c \) for the given balanced chemical equation:

\[ K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \]

From the given data:

\[ [N_2] = 0.048 \, M \]
\[ [O_2] = 0.093 \, M \]
\[ [N_2O] = 6.55 \times 10^{-21} \, M \]

Substitute these values into the expression for \( K_c \):

\[ K_c = \frac{(6.55 \times 10^{-21})^2}{(0.048)^2 \times (0.093)} \]

Calculate the numerator:

\[ (6.55 \times 10^{-21})^2 = 4.29 \times 10^{-41} \]

Calculate the denominator:

\[ (0.048)^2 \times 0.093 = 2.14 \times 10^{-4} \]

Finally, divide the numerator by the denominator to find \( K_c \):

\[ K_c = \frac{4.29 \times 10^{-41}}{2.14 \times 10^{-4}} = 2.0 \times 10^{-37} \]

Therefore, the correct answer is:

b. \( 2.0 \times 10^{-37}
Transcribed Image Text:--- **Chemical Equilibrium Constant Calculation** Consider the equilibrium reaction: \[ 2N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g) \] In a particular experiment, the equilibrium concentration of N₂ is 0.048 M, of O₂ is 0.093 M, and of N₂O is 6.55 x 10⁻²¹ M. What is the value of the equilibrium constant \( K_c \)? **Possible answers**: a. \( 1.5 \times 10^{-18} \) b. \( 2.0 \times 10^{-37} \) c. \( 2.2 \times 10^{-36} \) d. \( 3.1 \times 10^{-17} \) e. \( 5.0 \times 10^{36} \) --- Interpretation: To determine the equilibrium constant \( K_c \), use the expression for the equilibrium constant \( K_c \) for the given balanced chemical equation: \[ K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \] From the given data: \[ [N_2] = 0.048 \, M \] \[ [O_2] = 0.093 \, M \] \[ [N_2O] = 6.55 \times 10^{-21} \, M \] Substitute these values into the expression for \( K_c \): \[ K_c = \frac{(6.55 \times 10^{-21})^2}{(0.048)^2 \times (0.093)} \] Calculate the numerator: \[ (6.55 \times 10^{-21})^2 = 4.29 \times 10^{-41} \] Calculate the denominator: \[ (0.048)^2 \times 0.093 = 2.14 \times 10^{-4} \] Finally, divide the numerator by the denominator to find \( K_c \): \[ K_c = \frac{4.29 \times 10^{-41}}{2.14 \times 10^{-4}} = 2.0 \times 10^{-37} \] Therefore, the correct answer is: b. \( 2.0 \times 10^{-37}
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